Let $f(\mathrm{x})= \begin{cases}(1+\mathrm{ax})^{1 / \mathrm{x}} & , \quad \mathrm{x} \lt 0 \\ 1+\mathrm{b} & , \quad \mathrm{x}=0 \\ \frac{(\mathrm{x}+4)^{1 / 2}-2}{(\mathrm{x}+\mathrm{c})^{1 / 3}-2} & ,\end{cases}$
be continuous at $x=0$. Then $e^a b c$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & f\left(0^{-}ight)=\mathrm{e}^{\lim _{x ightarrow 0} \frac{\mathrm{ax}}{\mathrm{x}}}=\mathrm{e}^{\mathrm{a}} \ & \mathrm{f}(0)=1+\mathrm{b} \ & \mathrm{f}\left(0^{+}ight)=\frac{\frac{1}{2 \sqrt{\mathrm{x}+4}}}{\frac{1}{3}(\mathrm{x}+\mathrm{c})^{-\frac{2}{3}}}=\frac{\frac{1}{2(2)}}{\frac{1}{3} \cdot \mathrm{c}^{-\frac{2}{3}}} \ & =\frac{3}{4} \mathrm{c}^{2 / 3} \end{aligned}$ Also at $\mathrm{x}=0$; $c^{1 / 3}=2 \Rightarrow c=8$ So $\mathrm{f}\left(0^{+}ight)=\frac{3}{4}(8)^{2 / 3}=3$ Now, $\mathrm{e}^{\mathrm{a}}=\mathrm{b}+1=3$ $\mathrm{e}^{\mathrm{a}} . \mathrm{b} \cdot \mathrm{c}=3 \cdot 2 \cdot 8=48$

Let $f(\mathrm{x})= \begin{cases}(1+\mathrm{ax})^{1 / \mathrm{x}} & , \quad \mathrm{x} \lt 0 \\ 1+\mathrm{b}…

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